This is a tad hard for G1 (At MOP, most of Blue was trolled by this problem.) First, use the fact that \[ 90^{\circ} + \frac{1}{2} \angle A = \angle B'OC' > \angle BOC = 2 \angle A \]to obtain $\angle A < 60^{\circ}$. Now $M$ be the midpoint of $BC$. Then \[ OL \ge OM = R\cos A > R/2 \]so we are done.

Case 1: $\angle A< 60^\circ$: FLet $r=LB'=LC', R=OB=OC$. Note that Stewart's on $\triangle OBC$ and point $L$ gives $R^2-r^2=\frac{a^2bc}{(b+c)^2}\le \frac{a^2}{4}$. Since $R=\frac{a}{2\sin A}> \frac{a}{\sqrt{3}}$, we have $3R^2> a^2\implies R^2-r^2<\frac{3R^2}{4}\implies R< 2r$, which means (since the circle with center $L$ passes through $O$) that the circles must intersect twice, as desired. Case 2: $\angle A\ge 60^\circ$. Note that we have: \[120^\circ\ge\angle B+\angle C=180^\circ-\angle B'LB-\angle CLC'= \angle B'LC'=360^\circ-2(\angle LB'O+\angle LC'O)= 2((90^\circ-\angle LB'O)+(90^\circ-\angle LC'O))= 2(\angle OB'A+\angle OC'A)>2(\angle OBA+\angle OCA)= 2((90^\circ-\angle B)+(90^\circ-\angle C))=2\angle A\ge 120^\circ\] Which is a contradiction, since this implies $120^\circ>120^\circ$. So this case is impossible. This covers all cases, so we're done.

Let $H$ be the orthocenter of $\triangle ABC$.It is well-known that $\angle BHC=180^{\circ}-\angle A$.So the reflection$(H')$ of $H$ in $BC$ lies on $\odot ABC$. Let $O'$ be the reflection of $O$ in $BC$.If $O'$ is outside $\odot ABC$,then $\odot ABC,\omega$ intersect at two points. But then $\angle BO'C<\angle BH'C$ $\Rightarrow \angle BOC<\angle BHC$ $\Rightarrow 2\angle A<180^{\circ}-\angle A$ $\Rightarrow \angle A<60^{\circ}$. So it suffices to prove that $\angle A<60^{\circ}$. Since $\angle B,\angle C$ are acute and $LB' \bot AB,LC' \bot AC$;$B',C'$ lie on the segments $AB,AC$ respectively. So $\angle BOC<\angle B'OC'$ $\Rightarrow 2\angle A<\frac {1}{2}(360^{\circ}-\angle B'LC')$ $=180^{\circ}-\frac {1}{2}(180^{\circ}-\angle A)$[The last arguement follows from the fact that $AB'LC'$ is cyclic.] $=90^{\circ}+\frac {1}{2}\angle A$ $\Rightarrow A<60^{\circ}$ So the two circles intersect at two points.

Let point I is the incenter of the triangle, We can think two cases; if I=O we can find the solution easily; otherwise just think triangle ALO and use Stewart's theorm that we can find the claim R<2r' easily. we use that things; 1. point I is the intersection of AL and omega 2. IO^2=R^2-2Rr 3. r'=r*(a+b+c)/(b+c) I think it's not so hard to find;

Has anybody tried solving this problem trigonometricaly? The sine theorem tells us that $R=\frac{a}{2 \sin{\beta + \gamma}}$ And one can easily compute that $r= \frac{a\sin{\beta}\sin{\gamma}}{\sin{\beta}+\sin{\gamma}}$ The problem then reduces to proving the following inequality: $\sin{\beta}+\sin{\gamma} < 4\sin{\beta}\sin{\gamma}\sin{\beta + \gamma}$ for suitably constrained $\beta,\gamma$. I'm stuck here. Can anybody help out?

I got something weird...

I'm sure something is wrong. Did I read the question wrongly? EDIT: Oops, why did I type X? It's supposed to be L.

meowme wrote: Therefore, $X$ lies in the circumcircle (obviously), and in circle $\omega$ (even more obviously). What is $X$?

Sorry, I meant L.

meowme wrote: Hence, there are more than 2 common points in the interiors of the 2 circles. OK, I finally noticed this now. The problem is asking about the intersection of two circles, not two disks. In other words, you need the circumferences of the circles to intersect twice.

I might be being very silly, but doesn't 2 common points in the interior also imply that the circumferences intersect twice?

Nope. [asy][asy] draw(CR((0,0), 2012)); draw(CR((256, 42), 1337)); [/asy][/asy]

Ah yes. Is that an easy fix, though? I'm not sure.

EDIT: Actually, it must be true, otherwise the question would be false. I think that we can note that the triangle is acute. It reduces to $2a^2b^2 + 2a^2c^2 + 2b^2c^2 \ge abc(b+c) + a^4 + b^4 + c^4$. Any ideas on this?

I know I'm missing something but I found something weird $L$ is the foot of the angle bisector from $A$ Since $AB'C'$ is isosceles, Both $A,L$ lie on the perpendicular bisector of $B'C'$ $\implies L$ lies on the angle bisector of $\angle A$

utkarshgupta wrote: Since $AB'C'$ is isosceles, Both $A,L$ lie on the perpendicular bisector of $B'C'$ $\implies L$ lies on the angle bisector of $\angle A$ This is all correct but it doesn't solve the problem.

Ok I agree Here's my proof which I think works Let the radius of $\omega$ be $r$ and that of circumcircle be $R$ We need to prove that $R<2r$ CASE 1 $\angle A \ge 60$ Now quite obviously $AO<AL<2LB'<2r$ CASE 2 $\angle A <60 $ Now observe $r(cosec B + cosec C)=2R \sin A$ Using Jensen's and $A<60$, we get $2r cosec 60 \ge 2r cosec (90-\frac{A}{2}) \ge r( cosec A+cosec B) = 2R \sin A > \sqrt{3} R$ $\implies 2r >R$ Thus the result

It suffices to show $OL > \dfrac{R}{2}$ which can be done through $OL \ge OM > \dfrac{R}{2}\implies \angle A < 60^{\circ}$. Suppose $\angle A \ge 60^{\circ}$; then $\angle BOC > 120^{\circ}$ while $\angle B'OC' = 180^{\circ}-\dfrac{1}{2}\angle B'LC' = 120^{\circ}$. However, clearly $\angle BOC < \angle B'OC'$ contradiction. Thus $\angle A < 60^{\circ}$ and we're done.

I liked this a lot! Few years ago, I bashed it out to get a nasty inequality which somehow worked. I will post a synthetic solution now : Let $D$ be the midpoint of $BC$ and $R$ be the circumradius of $\triangle ABC$. Note that $AL$ bisects angle $BAC$ and $O$ lies inside $\triangle ABC$ so $$\angle B'OC'>\angle BOC \Longrightarrow 90^{\circ}+\frac{A}{2}>2A \Longrightarrow A<60^{\circ}.$$Extend ray $OL$ beyond $L$ to meet the circumcircle $\gamma$ of $\triangle ABC$ at $X$. Evidently, $$OL \ge OD=R\cos A>\frac{R}{2} \Longrightarrow OL>LX,$$so $\omega$ contains point both in the interior and the exterior of $\gamma$. It follows that $\omega$ and $\gamma$ have two points in common. $\square$

Let $R$ be circumradius on $(ABC)$ and $R'$ be radius of circle with center $L$. We have $R' \geq R \cos A$. This reduces to proving $\angle A < 60^{\circ}$ We have $\angle B'OC' > \angle BOC$. This gives $180^{\circ}-A > 2A \implies 60>A$ and hence we are done.

Weird problem, but easy if you figure out the right thing to prove. $\angle B'OC' > \angle BOC$ gives $\angle BAC < \pi/3$. Hence, if $X$ is the antipode of $O$ in $\omega$, then $OX=2OL \geq R \cos A > R$, since $R \cos A$ is the length of the perpendicular from $O$ to $BC$ and hence the minimum distance between $O$ and any point on $BC$. Hence, a point of $\omega$ lies strictly inside, and a point of $\omega$ lies strictly outside $(ABC)$, and so they must intersect at 2 points $\blacksquare$

Absolutely magnificent problem. WakeUp wrote: Let $ABC$ be an acute triangle. Let $\omega$ be a circle whose centre $L$ lies on the side $BC$. Suppose that $\omega$ is tangent to $AB$ at $B'$ and $AC$ at $C'$. Suppose also that the circumcentre $O$ of triangle $ABC$ lies on the shorter arc $B'C'$ of $\omega$. Prove that the circumcircle of $ABC$ and $\omega$ meet at two points. Proposed by Härmel Nestra, Estonia Firstly, by an obvious homothety, it follows that $L \overset{\text{Def}}{=} \overline{AI} \cap \overline{BC} .$ Also, using the condition that $O$ lies on the minor arc $B^{\prime}C^{\prime}$, it is consequent that $\angle BAC < 60^{\circ}$. Now, randomly pick a point $X$ on $\omega$ and denote by $\S(OX)$ the length of segment $\overline{OX}$. If $O^{\prime}$ is the anti-pode of $O$ with respect to $\omega$, then we have $\S(OO^{\prime}) = 2 \cdot \S(OL) > 2 \cdot \S(OM) =2R \cdot \cos{\angle BAC}$, where $M$ is the midpoint of $\overline{BC}$. Since $\angle BAC < 60^{\circ}$ it follows that $2R \cdot \cos{\angle BAC} > R$. Also since $B^{\prime}$ lies on segment $\overline{AB}$ we have $\S(OB^{\prime}) < \S(OB) = R$. So let $\S(OB^{\prime}) = R-a_1 $ and $\S(OO^{\prime}) = R + a_1$ for some $a_1,a_2 \in \mathbb{R^+}$. Thus, as $X$ moves on the minor arc $B^{\prime}O^{\prime}$, $\S(OX)$ spans over all values in the interval $[R - a_{1} , R + a_{2}] \implies$ there exists a point $P$ on the minor arc $\widehat{B^{\prime}O^{\prime}}$ such that $\S(OP) = R \implies P \in \odot(ABC)$. Similarly there exists a point on the minor arc $\widehat{C^{\prime}O^{\prime}}$. Thus, $\omega$ meets $\odot(ABC)$ in exactly $2$ points.

Can someone check? Let the perpendicular to $BC$ through $L$ intersect $(ABC)$ at $X$ and $\omega$ at $Y$. $LX<OX-OL$ from triangle inequality. $OX=OC=OB$ so $LX<OC-OL \leq OC$. $OC \leq OL = LY$ because $Y,O$ lie on $\gamma$. Hence $LX < LY$ and hence the circles intersect at two points

WakeUp wrote: Let $ABC$ be an acute triangle. Let $\omega$ be a circle whose center $L$ lies on the side $BC$. Suppose that $\omega$ is tangent to $AB$ at $B'$ and $AC$ at $C'$. Suppose also that the circumcentre $O$ of triangle $ABC$ lies on the shorter arc $B'C'$ of $\omega$. Prove that the circumcircle of $ABC$ and $\omega$ meet at two points. Proposed by Härmel Nestra, Estonia Kinda difficult for a G1 We see that $L$ is the foot of the angle bisector. We see that $\angle B'OC' = \frac{360 - \angle B'LC'}{2} = \frac{180 - \angle BAC}{2} > \angle BOC = 2 \angle BAC \implies \angle BAC < 60^\circ$. Now, we see that $OL \geq d(O, \overline{BC}) = OA \cos (\angle BAC) > \dfrac{OA}{2}$, which means that $\omega$ and circumcircle of $\triangle ABC$ will intersect twice (because any circle with radius $\dfrac{OA}{2}$ passing through $O$ will be tangent to circumcircle of $\triangle ABC$) as desired.

this problem I think is very hard for the G1.

All angles are in degrees. Let $R$ be the circumradius of $ABC$, and let $\Gamma$ be the circumcircle. Notice that $$2\angle A = \angle BOC < \angle B'OC' = 180-\frac{1}{2}\angle B'LC' = 180-\frac{1}{2}(180-\angle A)$$and therefore $\angle A < 60 \Rightarrow \angle BOC < 120$. However, letting $M$ be the midpoint of $BC$, we have $OL > OM > OB\sin(30)=\frac{1}{2}R$. Therefore, letting $O'$ be the reflection of $O$ about $L$ (clearly lying on $\omega$), $O'$ must be outside $\Gamma$. Since there exist points on $\omega$ that are outside $\Gamma$ and there exist points outside $\Gamma$, $\Gamma$ and $\omega$ must intersect at exactly two points.

$\angle BOC = 2\angle A$ and $\angle B'OC' = \angle A + \angle OB'A + \angle OC'A = \angle A + \frac{\angle B'LC'}{2} = \frac{\angle A}{2} + 90$ so we have $\frac{\angle A}{2} + 90 > 2\angle A \implies \angle A < 60$. Let $M$ be midpoint of $BC$ Note that $OM = R\cos A > \frac{R}{2}$ and $OL \ge OM$ so radius of $\omega$ is bigger than radius of $ABC$ so they meet at $2$ points.

Let the foot of perpendicular from $O$ to $\overline{BC}$ be $D$, note that points $B', C'$ are on segments $|AB|, |AC|$, respectively. So $$2\angle{BAC}=\angle{BOC}>180-\frac{1}{2}\angle{B'LC'}=180-\frac{1}{2}(180-\angle{BAC})=90+\frac{1}{2}\angle{BAC}\Rightarrow \angle{BAC}<60\Rightarrow \angle{OCD}>30$$thus $$\frac{OL}{OC}\geq \frac{OD}{OC}=\sin{OCD}>\sin{30}=\frac{1}{2}\Rightarrow 2|OL|>|OC|$$taking into account that function $f(x)=\sin{x}$ in increasing on interval $[0,\frac{\pi}{2}]$, therefore $(B'OC')$ must have a point outside of $(ABC)$ and since it already has points inside, it intersects $(ABC)$ at two points.

-_- It suffices to show $2OL > BO.$ By sine law, $\frac{OB}{OL}=\frac{\sin(\angle OLB)}{\sin(OBL)}<\frac{1}{\sin \angle OBC}.$ It remains to show that $\angle OBC>30^\circ.$ Let $\angle OBC=\alpha$ then $\angle BOC=180^\circ-2\alpha$ and $\angle BAC = 90^\circ-\alpha.$ We have $\angle B'LC'=90^\circ+\alpha$ so $\angle B'OC'=135^\circ-\frac{\alpha}{2}.$ Obviously $\angle B'OC'>\angle BOC$ implies result.

Literally the same as #4 oops

It is sufficient to prove that $R \leq 2 OL$. The key claim is that Claim: $\angle BAC \leq 60^{\circ}$. Proof: We know that $\angle B'OC' < \angle BOC$. Notice that $\angle B'OC' = \angle B'OL + \angle C'OL = \angle OB'L + \angle OC'L$, so \begin{align*}\angle B'OC' &= \frac{360^{\circ} - \angle B'LC'}{2}\\ &= \frac{360^{\circ} - (180^{\circ} - \angle BAC)}{2}\\ &= \frac{\angle BAC + 180^{\circ}}{2} < 2\angle BAC \\ \implies \angle BAC &\leq 60^{\circ}.\end{align*} Therefore, $\angle BOC \leq 120^{\circ} \implies \angle OBC \leq 30^{\circ} \implies OB \leq 2 OM \implies R \leq 2 OM \leq 2 OL$ and we're done. $\square$

Let $r$ be the radius of $\omega$, and $R$ be the circumradius of the triangle. The problem reduces to showing that $2r > R$. We separate this into two cases: - $\angle A \geq 60^\circ$: Let $K$ and $M$ be the tangency points of $\omega$ onto $\overline{AC}$ and $\overline{AB}$. We know that $\tan \angle LAK \geq \frac{1}{\sqrt{3}}$, so $AK \leq r\sqrt{3}$. Since $O$ lies on the interior of the triangle, it follows that $R = AO < AK \leq r\sqrt{3}$. Thus, $R < 2r$ in this case. - $\angle A < 60^\circ$: Then, $\angle B + \angle C > 120^\circ$. Use Stewart's on $\triangle BOC$ with cevian $\overline{OL}$. It follows that $$r^3\left(\frac{1}{\sin \angle B \sin \angle C}\right)\left(\frac{1}{\sin \angle B} + \frac{1}{\sin \angle C}\right) + r^3\left(\frac{1}{\sin \angle B} + \frac{1}{\sin \angle C}\right) = R^2r\left(\frac{1}{\sin \angle B} + \frac{1}{\sin \angle C}\right),$$or $$r^2\left(\frac{1}{\sin \angle B \sin \angle C} + 1\right) = R^2.$$However, $\sin \angle B \sin \angle C = \frac12(\cos (\angle B - \angle C) - \cos(\angle B + \angle C))$. Also, $\cos(\angle B - \angle C) > \cos(60^\circ) = \frac12$, while $\cos (\angle B + \angle C) < -\frac12.$ Thus, $\sin \angle B \sin \angle C > \frac12$, and $\frac{1}{\sin\angle B \sin\angle C} + 1 < 3$. Now, $R^2 < 3r^2$ so $R < 2r$, proving the result. I didn't realize the second case is impossible, but I think it still works?

The conclusion is equivalent to $$R_{(ABC)}-R_\omega<OL<R_{(ABC)}+R_\omega.$$Since $R_\omega=OL$, the second half of the inequality is trivial. Thus, it remains to prove that $$OL>\frac{1}{2}R.$$ Let $M$ be the midpoint of $BC$. Clearly, $OM\leq OL$, so it suffices to show that $OM>\frac{1}{2}R$. This is equivalent to $\angle OBM>30,$ or $\angle BAC<60.$ Therefore, it suffices to show that $\angle BAC<60$. Let $\angle BAC=2\alpha$. Then, $$\angle BOC=4\alpha<\angle B'OC'=\frac{1}{2}(360-\angle B'LC')=\frac{1}{2}(360-(180-2\alpha))=90+\alpha.$$Solving this inequality for $\alpha$, we have $\alpha<30$, so we are done.

wow this is surprisingly nontrivial for g1 - really nice, would never have thought to look at angle B'OC' until i saw the hint in egmo Clearly it suffices to show that $OL \leq R/2$. However, $OL$ is greater than the distance from $O$ to $BC$, which is $R \cdot \cos(A)$, hence we must show $\cos(A) \leq \frac{1}{2}$, or that $\angle A \leq 60^\circ .$ Note that \[\angle B'OC' = 90^\circ + \frac{\angle A}{2} \geq 2\angle A = \angle BOC \implies \angle A \leq 60^\circ\]as desired.

surprised i'm the only one who completely trigbashed this Note that if $\angle A \ge 90$, $O$ lies on the side of $BC$ not containing $A$ while the minor arc in question obviously is on the side of $BC$ containing $A$, contradiction (note that when $\angle A =90$ we need the stronger statement that the minor arc doesn't intersect $BC$ which is obvious). Thus $\angle A < 90$. Note that $BL=\frac{ac}{b+c}$ by the angle bisector theorem, and so the radius of $\omega$ is $\frac{ac \sin B}{b+c} = \frac{2K}{b+c}$ where $K$ is the area of $ABC$. Note that $OL \ge R\cos A$ since $R\cos A$ is the length of the altitude in $OBC$. Thus if we prove the radius of $\omega$ is bigger than $R-R\cos A$, then we can find a point on $\omega$ that is outside $(ABC)$ and then the conclusion obviously follows (this point is the antipode of $O$). Note that $$AB' = b - CB' = b - \frac{ab}{b+c} \cdot \frac{a^2+b^2-c^2}{2ab} = b-\frac{a^2+b^2-c^2}{2(b+c}) = \frac{(b+c)^2-a^2}{2(b+c)}.$$Also since $O$ is on the minor arc, $R=AO \le AB' = \frac{(b+c)^2-a^2}{2(b+c)}$. Now $$R-R\cos A \le \frac{(b+c)^2-a^2}{2(b+c)} \cdot \frac{a^2-(b-c)^2}{2bc} = \frac{4K^2}{bc(b+c)}$$. Now we prove $\frac{4K^2}{bc(b+c)} \le \frac{2K}{b+c} \equiv 2K = bc \sin A \le bc$, which is obviously true and we're done.

As $L$ is equidistant from $AC$ and $AB$, it lies on the $A$-angle bisector. From now on use vectors with $\vec{O} = 0$. Hence $L = \tfrac{b\vec{B} + c\vec{C}}{b+c}$. Note that it suffices to show that $|\vec{L}| > \tfrac{1}{2}R$. Compute \begin{align*} |\vec{L}|^2 = \vec{L}\cdot \vec{L} &= \frac{1}{(b+c)^2}(b^2\vec{B}\cdot \vec{B} + c^2\vec{C}\cdot \vec{C} + 2bc\vec{B}\cdot \vec{C})\\ &= \frac{b^2 + c^2 + 2bc\cos \angle BOC}{(b+c)^2}\cdot R^2\\ &= \frac{b^2 + c^2 + 2bc\cos 2A}{(b+c)^2}R^2. \end{align*}We can bound $\cos 2A$ by observing that \[ 2A = \angle BOC < \angle B’OC’ = 180^\circ - \frac{1}{2}\angle B’LC’ = 180^\circ - \frac{1}{2}(180^\circ - A) = 90^\circ + \frac{A}{2},\]meaning $A < 60^\circ$, giving $\cos 2A > -\tfrac{1}{2}$. Hence, \[ |\vec{L}|^2 > \frac{b^2 + c^2 - bc}{(b+c)^2}R^2 = R^2\left(1 - \frac{3bc}{(b+c)^2}\right).\]To show $|\vec{L}|^2 > \tfrac{R^2}{4}$, it suffices to show $1 - \tfrac{3bc}{(b+c)^2}\ge \tfrac{1}{4}$. But this is obviously true, so we are done.

I find it interesting that despite all of the solutions here, I don't think there is a SINGLE diagram. AFTSOC they were tangent; we would have OB=2OL, so we'll prove 2<\frac{OB}{OL}=\frac{\sin{OLB}}{\sin{OBL}}\le\frac1{\sin{OBL}}; we can prove OBC>30 which is equivalent to proving BAC<60. Indeed, B'LC'=180-A\implies 90+\frac A2=B'OC'>BOC=2A\implies A<60, which finishes. However, this implies sinOBL>1/2 so 1/sinOBL<2, which is wrong direction inequality, can someone fix? https://paste.pics/c5691cf0f05ab5131deb94e4f2a82ff0

In order for the condition to hold, we need \[90 + \frac A2 = \angle B'OC' < \angle BOC = 2A \implies A < 60 \implies \angle OBC > 30.\] It suffices to show the reflection of $O$ about $L$ lies outside the circumcircle, or $OL > \frac R2$, which holds as \[OL \ge d(O, BC) = R \cdot \sin \angle OBC > \frac R2. \quad \blacksquare\]

Cute question

Bleh. We observe \[ 90^\circ + A/2 = \angle B'OC' > \angle BOC = 2A \], whence $A < 60^\circ.$ Thus the reflectiom of $O$ over $BC$ lies outside the circumcircle, implying the result.

Let $O'$ be the reflection of $O$ across $L$. Note that the conclusion is equivalent to showing $OO' > OB$. Which is equivalent to showing $\angle OBO' > \angle BOO'.$ Indeed this is equivalent to showing \[2 \angle OBC > \angle BOK \iff 180 - \angle BOC > 90 - \angle OBC = \angle BOC/2 \iff \angle BOC > 120.\] Which is true as, \[\angle BOC > \angle B'OC' = 90 + A/2 = 90 + \angle BOC/4 \iff \angle BOC > 120\] as needed.

Notice that $180 - \angle A = \angle B'LC' = 2 (180 - \angle B'OC') > 2 (180 - \angle BOC) = 360 - 4 \angle BAC $, so we have $\angle BAC > 60$, implying that the reflection of $O$ over the midpoint of $BC$ lies outside of $\omega$, thus we see the reflection of $O$ over $L$ also lies outside of $\omega$, done.

Note that $OL \geq OM = R \cos \frac{\angle A}{2},$ so it suffices to show that $\angle A \leq 60^\circ.$ To do this, observe that $90^\circ + \frac{\angle A}{2} = \angle B'OC' \geq \angle BOC = 2 \angle A$ since $B', C'$ are within segments $AB, AC$ respectively. This equation yields the desired result, so we are done. QED